It’s been a while since my last math post. It seems that my previous assertion correlating the quality of question with the period of final exams could not have been further from the truth. It now seems quite likely that the occurrence of a few “good” questions on Yahoo Answers in a short period of time a few weeks back was more of a fluke rather than the norm.

**Problem:** Find the standard matrix for the linear transformation which reflects points in the x-y plane across the line y = \frac{-2x}{3}.

**Solution:** To find the matrix representing a given linear transformation all we need to do is to figure out where the basis vectors, i.e.

\vec{e_{1}}=\begin{bmatrix} 1\\ 0 \end{bmatrix} and \vec{e_{2}}=\begin{bmatrix} 0\\ 1 \end{bmatrix}

are mapped under the transformation.

This is easy to see from the fact that since \vec{e_{1}} and \vec{e_{2}} form a basis of \mathbb{R}^{2} any element

\vec{u}\ \epsilon \ \mathbb{R}^{2}

can be written as

\vec{u} = a \cdot \vec{e_{1}} + b \cdot \vec{e_{2}}

for some a,b\ \epsilon\ \mathbb{R}.

Now let

A : \mathbb{R}^{2} \to \mathbb{R}^{2}

be a linear transformation. Then, by definition

A(\vec{u}) = A(a \cdot \vec{e_{1}} + b \cdot \vec{e_{2}}) = a \cdot A( \vec{e_{1}} ) + b \cdot A( \vec{e_{2}} )

Thus, the image of any element \vec{u}\ \epsilon \ \mathbb{R}^{2} under a linear transformation is completely determined by the image of the basis of the transformation’s domain.

Let’s find the image of \vec{e_{1}}=\begin{bmatrix} 1\\ 0 \end{bmatrix} under a reflection across the line given by y = \frac{-2x}{3}. First, we need to find a line perpendicular to y = \frac{-2x}{3} that passed through the point p_{1} = (1, 0).

In general, a line perpendicular to y = mx + b is given by y = \frac{-x}{m}, i.e in our case, y = \frac{3x}{2}. To …